Focal length of lens at border of two liquids

AI Thread Summary
The discussion revolves around calculating the optical power of a lens situated at the interface of two liquids using the spherical diopter equation. The user attempts to derive the focal length but finds discrepancies between their solution and the textbook answer, which employs ray path and geometric methods. They explore the possibility of treating the lens surfaces separately and reference the lensmaker's formula to address their confusion. The user notes that their approach yields two focal lengths, while the textbook solution does not. Ultimately, the conversation highlights the complexities of lens optics in different refractive media.
matej1408
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Homework Statement


I need to find optical power (reciprocal focal length) of this system with thin lens
Capture.png


Homework Equations


I tried to solve this using spherical diopter equation
n1/a+n2/b=(n2-n1)/R
where a is object distance and b is image distance

The Attempt at a Solution


equation for first diopter
n1/a+n/b'=(n-n1)/R
equation for second diopter
-n/b'+n2/b=-(n2-n)/R
adding these two equation i have
n1/a+n2/b=(2*n-n1-n2)/R
putting a→∞, b should be focal length then
so:
1/f=1/b=(2*n-n1-n2)/(n2*R)
but textbook solution is:
Capture.png

where -x=a and x'=b
i figured out this solution doesn't have two different focal length solutions (one from each side) and my solution have that. They solved problem using ray path and geometry.
I'm wondering what is wrong with my solution


 
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What might work is treating the two surfaces of the lens separately. That is the first produces an image which becomes the object for the second. Start out with an object far away - sorry, that is what your solution does in effect.
 
Last edited:
Look at the derivation of the lensmaker's formula and modify it to the case where n1 ≠ 1.
 
rude man said:
Look at the derivation of the lensmaker's formula and modify it to the case where n1 ≠ 1.
in my textbook is solved by derivation of lensmaker's formula, but i want to know why my solution isn't correct
 
The power of the lens seems to be the sum of the powers of the two surfaces, which would then be
P = Pleft + Pright
= (n - n1)/R + (n2 - n)/-R
= 1/R (2n - n1 - n2)
I am referencing Jenkins and White - Fundamentals of Optics.
Each of the surfaces of the lens has two associated focal lengths.
So who knows what the f in the formula in the book refers to!
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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